Question

Use Theorem 2.1.1 to verify the logical equivalences in 50–54. Supply a reason for each step.

∼(p ∨ ∼q) ∨ (∼p ∧ ∼q) ≡ ∼p

Answer 1

To verify the logical equivalence ∼(p ∨ ∼q) ∨ (∼p ∧ ∼q) ≡ ∼p using Theorem 2.1.1, we can simplify both sides of the equation and show that they are equivalent.

Step-by-step explanation:

To verify the logical equivalence ∼(p ∨ ∼q) ∨ (∼p ∧ ∼q) ≡ ∼p using Theorem 2.1.1, we can simplify both sides of the equation and show that they are equivalent.

Starting with the left-hand side (∼(p ∨ ∼q) ∨ (∼p ∧ ∼q)), we can use De Morgan's Laws to distribute the negation over the parentheses:

1. ∼(p ∨ ∼q) ∨ (∼p ∧ ∼q)
2. (∼p ∧ q) ∨ (∼p ∧ ∼q)
3. (∼p ∧ (q ∨ ∼q))
4. (∼p ∧ T)
5. ∼p

Now, let's analyze the steps:

• In step 1, we apply De Morgan's Law ∼(p ∨ ∼q) = (∼p ∧ q).
• In step 2, we use the distributive property on (∼p ∧ q) ∨ (∼p ∧ ∼q) to get (∼p ∧ (q ∨ ∼q)).
• In step 3, we apply the law of excluded middle (q ∨ ∼q = T).
• In step 4, we use the identity law (∼p ∧ T = ∼p).
• Finally, in step 5, we arrive at ∼p, which matches the right-hand side of the equation.

Therefore, the logical equivalence ∼(p ∨ ∼q) ∨ (∼p ∧ ∼q) ≡ ∼p is verified using Theorem 2.1.1 and the step-by-step simplification process.